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College Physics Student Solutions Manual Chapter 28
201
CHAPTER 28: SPECIAL RELATIVITY 28.2 SIMULTANEITY AND TIME DILATION
1. (a) What is γ if cv 250.0= ? (b) If cv 500.0= ?
Solution (a) Using the definition of γ , where cv 250.0= :
0328.1)250.0(112/1
2
22/1
2
2
=⎥⎦
⎤⎢⎣
⎡−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=
−−
cc
cv
γ
(b) Again using the definition of γ , now where cv 500.0= :
15.11547.1)500.0(12/1
2
2
==⎥⎦
⎤⎢⎣
⎡−=
−
cc
γ
Note that γ is unitless, and the results are reported to three digits to show the difference from 1 in each case.
6. A neutron lives 900 s when at rest relative to an observer. How fast is the neutron moving relative to an observer who measures its life span to be 2065 s?
Solution Using 0tt Δ=Δ γ , where
2/1
2
2
1−
⎟⎟⎠
⎞⎜⎜⎝
⎛−=cv
γ , we see that:
.12944.2s 900s 2065
2/1
2
2
0
−
⎟⎟⎠
⎞⎜⎜⎝
⎛−===
Δ
Δ=
cv
tt
γ
Squaring the equation gives 22
21
2
222
vcc
cvc
−=⎟⎟
⎠
⎞⎜⎜⎝
⎛ −=
−
γ
Cross-‐multiplying gives 2
222
γcvc =− , and solving for speed finally gives:
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ccccccv 900.090003.0)2944.2(
11112/1
2
2/1
22
22 ==⎥
⎦
⎤⎢⎣
⎡−=⎟⎟
⎠
⎞⎜⎜⎝
⎛−=−=γγ
11. Unreasonable Results (a) Find the value of γ for the following situation. An Earth-‐bound observer measures 23.9 h to have passed while signals from a high-‐velocity space probe indicate that h 0.24 have passed on board. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
Solution (a) Using the equation ,0tt Δ=Δ γ we can solve for γ :
996.09958.0h 24.0h 23.9
0
===Δ
Δ=tt
γ
(b) γ cannot be less than 1.
(c) The earthbound observer must measure a longer time in the observer's rest frame than the ship bound observer. The assumption that time is longer in the moving ship is unreasonable.
28.3 LENGTH CONTRACTION
14. (a) How far does the muon in Example 28.1 travel according to the Earth-‐bound observer? (b) How far does it travel as viewed by an observer moving with it? Base your calculation on its velocity relative to the Earth and the time it lives (proper time). (c) Verify that these two distances are related through length contraction 3.20=γ .
Solution Using the values given in Example 28.1:
(a)
km 1.39m 101.386s) 1087.4)(m/s 10998.2)(950.0(
s) 1087.4)(950.0(368
60
=×=××=
×=Δ=−
−
cctvL
(b) km 0.433m 104.329)s 1052.1)(m/s 10998.2)(950.0( 2680 =×=××=Δ= −ctvL
(c) km 0.433m 104.333.20
m 101.386 23
0 =×=×
==γL
L
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28.4 RELATIVISTIC ADDITION OF VELOCITIES
22. If a spaceship is approaching the Earth at c100.0 and a message capsule is sent toward it at c100.0 relative to the Earth, what is the speed of the capsule relative to the ship?
Solution Using the equation )/(1 2cuv
uvuʹ′+
ʹ′+= , we can add the relativistic velocities:
[ ] cccc
cccuvuvu 198.0
/)100.0)(100.0(1100.0100.0
)/(1 22 =+
+=
ʹ′+
ʹ′+=
28. When a missile is shot from one spaceship towards another, it leaves the first at c950.0 and approaches the other at c750.0 . What is the relative velocity of the two
ships?
Solution We are given: cu 750.0= and cu 950.0=ʹ′ . We want to find v , starting with the
equation )/(1 2cuv
uvuʹ′+
ʹ′+= . First multiply both sides by the denominator:
uvcuuvu ʹ′+=ʹ′
+ 2 , then solving for v gives:
[ ] cccccc
cuuuuv 696.0
1/)950.0)(750.0(750.0950.0
1)/( 22 −=−
−=
−ʹ′−ʹ′
=
The velocity v is the speed measured by the second spaceship, so the minus sign indicates the ships are moving apart from each other (v is in the opposite direction as u ).
34. (a) All but the closest galaxies are receding from our own Milky Way Galaxy. If a galaxy ly 100.12 9× away is receding from us at c900.0 , at what velocity relative to us must we
send an exploratory probe to approach the other galaxy at c990.0 , as measured from that galaxy? (b) How long will it take the probe to reach the other galaxy as measured from the Earth? You may assume that the velocity of the other galaxy remains constant. (c) How long will it then take for a radio signal to be beamed back? (All of this is possible in principle, but not practical.)
Solution Note that all answers to this problem are reported to 5 significant figures, to distinguish the results.
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(a) We are given cv 900.0−= and cu 990.0= . Starting with the equation
)/(1 2cuvuvuʹ′+
ʹ′+= , we now want to solve for uʹ′ . First multiply both sides by the
denominator: uvcuvuu ʹ′+=ʹ′+ 2 ; then solving for the probe’s speed gives:
cccc
cccuvvuu 99947.0
]/)900.0)(990.0[(1)900.0(990.0
)/(1 22 =−−
−−=
−
−=ʹ′
(b) When the probe reaches the other galaxy, it will have traveled a distance (as seen on Earth) of 0 vtxd += because the galaxy is moving away from us. As seen from
Earth, cu 9995.0=ʹ′ , so .0
uvtx
udt
ʹ′+
=ʹ′
=
Now, vtxtu +=ʹ′ 0 and y 101.2064ly 1y) (1
0.9000.99947ly 1012.0 11
90 ×=×
−×
=−ʹ′
=c
ccvuxt
(c) The radio signal travels at the speed of light, so the return time tʹ′ is given by
,0
cvtx
cdt
+==ʹ′ assuming the signal is transmitted as soon as the probe reaches
the other galaxy. Using the numbers, we can determine the time:
y 101.2058y) 10.2064(0.900c)(1ly 101.20 111110
×=×+×
=ʹ′c
t
28.5 RELATIVISTIC MOMENTUM
39. What is the velocity of an electron that has a momentum of m/skg 1004.3 21 ⋅× − ? Note that you must calculate the velocity to at least four digits to see the difference from c .
Solution Beginning with the equation
( )[ ] 21221 cu
mumup−
== γ we can solve for the speed u .
First cross-‐multiply and square both sides, giving 22
2
2
2
1 upm
cu
=−
Then, solving for 2u gives ( ) ( ) ( )222
2
2222
//1/1
cpmp
cpmu
+=
+=
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Finally, taking the square root gives ( )222 / cpmpu
+= .
Taking the values for the mass of the electron and the speed of light to five significant figures gives:
( ) ( ) ( )[ ]m/s 102.988
kg 102.9979/m/skg 103.34kg 109.1094
m/skg 103.04
8
1/222821231
21
×=
⎭⎬⎫
⎩⎨⎧ ×⋅×+×
⋅×=
−−
−
u
28.6 RELATIVISTIC ENERGY
48. (a) Using data from Table 7.1, calculate the mass converted to energy by the fission of 1.00 kg of uranium. (b) What is the ratio of mass destroyed to the original mass,
mm /Δ ?
Solution (a) From Table 7.1, the energy released from the nuclear fission of 1.00 kg of uranium is J 108.0 13×=ΔE . So, 2
released0 mcEE Δ==Δ , we get
( )( )
g .890kg 108.89m/s 103.00
J 100.8 428
13
2 =×=×
×=
Δ=Δ⇒ −
cEm
(b) To calculate the ratio, simply divide by the original mass:
( )( )
10.98 108.89kg 1.00
kg 108.89 442
4−−
−
×=×=×
=Δ
mm
52. A π -‐meson is a particle that decays into a muon and a massless particle. The π -‐meson has a rest mass energy of 139.6 MeV, and the muon has a rest mass energy of 105.7 MeV. Suppose the π -‐meson is at rest and all of the missing mass goes into the muon’s kinetic energy. How fast will the muon move?
Solution Using the equation 2relKE mcΔ= , we can determine the kinetic energy of the muon
by determining the missing mass: 222rel )1()(KE cmcmmmc µµπ γ −=−=Δ=
Solving for γ will give us a way of calculating the speed of the muon. From the
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equation above, we see that:
.32072.1MeV 105.7MeV 139.61 ===
−−=+
−=
µ
π
µ
µµπ
µ
µπγmm
mmmm
mmm
Now, use ( )[ ] 21221 −
−= cvγ or ,11 22
2
γ=−
cv
so
cccv 0.6532(1.32072)
1111 22 =−=−=γ
58. Find the kinetic energy in MeV of a neutron with a measured life span of 2065 s, given its rest energy is 939.6 MeV, and rest life span is 900s.
Solution From Exercise 28.6, we know that γ = 2.2944, so we can determine the kinetic energy of the neutron: ( )( ) .MeV 1216MeV 939.612944.2)1(KE 2
rel =−=−= mcγ
64. (a) Calculate the energy released by the destruction of 1.00 kg of mass. (b) How many kilograms could be lifted to a 10.0 km height by this amount of energy?
Solution (a) Using the equation 2mcE = , we can calculate the rest mass energy of a 1.00 kg mass. This rest mass energy is the energy released by the destruction of that amount of mass:
( )( ) J 108.99J 108.988m/s 102.998kg 1.00 161682released ×=×=×== mcE
(b) Using the equation mgh=PE , we can determine how much mass can be raised to
a height of 10.0 km: ( )( ) kg 109.17m 1010.0m/s 9.80
J 108.99PE 1132
16
×=×
×==
ghm